I claim $n=6$ is the only answer. Suppose \[ n=\prod_{1\le i\le k}p_i^{\alpha_i}, \]where $p_1<\cdots<p_k$, $k\ge 1$, are primes, and assume first that $n$ is even. Then, by Euclid-Euler Theorem, $n$ is of form $2^{p-1}(2^p-1)$ with $2^p-1$ being a (Mersenne) prime. In this case, $\varphi(n)=2^{p-2}(2^p-2)$, hence $2^p-2=2^u$ for some $u$. For $u=1$, it is seen easily that $p=2$ works, yielding $n=6$. For $u\ge 2$, we obtain a contradiction via modulo 4. Now, assume $n$ is odd. Then, $n$ is square-free: $n=p_1\cdots p_k$ for some Fermat primes $p_1,\dots,p_k$. We then have \[ \sigma(n) = (1+p_1)(1+p_2)\cdots(1+p_k) = 2p_1p_2\cdots p_k. \]Comparing both sides through the lens of $v_2(\cdot)$, we find that $k=1$, that is $n$ is a prime. In this case, $\sigma(n)=n+1$, which clearly does not satisfy the condition. Hence, $n=6$ is indeed the only possibility.